Read the directions for making a toy parachute.

1. Cut thick string or twine into four pieces that are 9 inches long.

2. Cut a 12x5-inch rectangle out of a plastic garbage bag.

3. Using a hole punch, poke a hole into each corner of the rectangle.

4. Slide the end of each piece of string through a hole in the rectangle.

5. Tie a knot to the end of each string so that they are attached to the corners of the plastic rectangle.

6. Either tape or tie the loose ends of the string to a small object, such as an action figure or toy car.

7. Hold the top of the parachute so that the object hangs loosely below it.

8. Find a high place from which to drop your parachute, such as a chair or a small wall.

9. Softly toss your parachute into the air, and watch it float to the ground.

What would likely happen if someone skipped step 3?

The pieces of string would be different lengths.
The string would not attach to the object.
The string would not attach to the plastic.
The plastic rectangle would be a different size.

Answer:

Ceramics are generally made by taking mixtures of clay, earthen elements, powders, and water and shaping them into desired forms. Once the ceramic has been shaped, it is fired in a high temperature oven known as a kiln. Often, ceramics are covered in decorative, waterproof, paint-like substances known as glazes.

Explanation:

Answer:

made essentially from a nonmetallic mineral (such as clay) by firing at a high temperature.

Explanation:

Answer:

for 13

Explanation:

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Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

[tex]\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant[/tex]

Differentiating the above equation:

[tex]\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\[/tex]

[tex]\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\[/tex]

        [tex]= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}[/tex]

Answer:

silos to store grains

Explanation:

I got it from PLATO... hope this helps...

A broad range of subjects are included in the culinary arts. Food science and nutrition, ingredient quality, seasonality, flavors and textures, presentation and plate layout, and more are some of these. There are numerous specialties, sub-niches, and job titles that fall within this broad category.

Additionally, to help students become more well-rounded cooks, a variety of cooking techniques, numerous recipes, new ingredients, and various flavor profiles are presented to them.

The term “culinary arts” is used to refer broadly to the processes involved in preparing, cooking, plating, presenting, and serving food. It relates to meals and the goods, not sweets or breads. That make up a meal, such as appetizers, side dishes, and main courses.

Therefore, Students who pursue a culinary education have the opportunity to study under several of the industry's best chefs.

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Answer:

You cant update it if there isin´t any update available lol

Explanation:

Answer:

4

or

22

Explanation:

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is [tex]$\frac{3}{8}$[/tex]  in.

A) Determine the maximum shear stress developed in the shaft.

[tex]$\tau_{max}$[/tex] = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

[tex]$ \phi $[/tex] = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              [tex]$ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$[/tex]

                              [tex]$= 56.67 \pi \text{ rad/s}$[/tex]

Power [tex]$= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$[/tex]

          = 1424500 ft. lb/s

Torque, [tex]$T = \frac{P}{\omega}$[/tex]

                 [tex]$=\frac{1424500}{56.67 \pi}$[/tex]

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

[tex]$\tau_{max} = \frac{T_c}{J}$[/tex]

        [tex]$=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$[/tex]

      = 2.93 ksi

B). Angle of twist :

     [tex]$\phi = \frac{TL}{JG}$[/tex]

         [tex]$=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$[/tex]

         = 0.08002 rad

         = 4.58°

Answer: 150m

Explanation:

The following can be depicted from the question:

Dimensions of outer walls = 9.7m × 14.7m.

Thickness of the wall = 0.30 m

Therefore, the plinth area of the building will be:

= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)

= 10 × 15

= 150m

The plinth area will be 150m.

Given information

Dimensions of outer walls = 9.7m × 14.7m.

Thickness of the wall = 0.30 m

Plinth area of the building = (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)

Plinth area of the building = 10 × 15

Plinth area of the building = 150m

Therefore, the Option A is correct.

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